Why does ${\int}_{a}^{ab}\frac{1}{x}dx={\int}_{1}^{b}\frac{1}{t}dt$?

I can't understand how the integral having limits from $a$ to $ab$ in Step 1 is equivalent to the integral having limits from $1$ to $b$. I'm a beginner here. Please explain in detail.

$\begin{array}{rl}\mathrm{ln}(ab)={\int}_{1}^{ab}\frac{1}{x}dx& ={\int}_{1}^{a}\frac{1}{x}dx+{\int}_{a}^{ab}\frac{1}{x}dx\\ & ={\int}_{1}^{a}\frac{1}{x}dx+{\int}_{1}^{b}\frac{1}{at}d(at)\\ & ={\int}_{1}^{a}\frac{1}{x}dx+{\int}_{1}^{b}\frac{1}{t}dt\\ & =\mathrm{ln}(a)+\mathrm{ln}(b).\end{array}$

I can't understand how the integral having limits from $a$ to $ab$ in Step 1 is equivalent to the integral having limits from $1$ to $b$. I'm a beginner here. Please explain in detail.

$\begin{array}{rl}\mathrm{ln}(ab)={\int}_{1}^{ab}\frac{1}{x}dx& ={\int}_{1}^{a}\frac{1}{x}dx+{\int}_{a}^{ab}\frac{1}{x}dx\\ & ={\int}_{1}^{a}\frac{1}{x}dx+{\int}_{1}^{b}\frac{1}{at}d(at)\\ & ={\int}_{1}^{a}\frac{1}{x}dx+{\int}_{1}^{b}\frac{1}{t}dt\\ & =\mathrm{ln}(a)+\mathrm{ln}(b).\end{array}$